Tính một cách hợp lí:
\(A = \left( { - \dfrac{3}{{11}}} \right) + \dfrac{{11}}{8} - \dfrac{3}{8} + \left( { - \dfrac{8}{{11}}} \right)\)
K
Khách
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Những câu hỏi liên quan
9 tháng 3 2022
d: \(=\dfrac{-7}{9}\left(\dfrac{3}{11}+\dfrac{8}{11}\right)+1+\dfrac{7}{9}=1\)
e: \(=\dfrac{1}{5}\left(\dfrac{10}{19}+\dfrac{9}{19}\right)-\dfrac{2}{35}=\dfrac{1}{5}-\dfrac{2}{35}=\dfrac{5}{35}=\dfrac{1}{7}\)
f: \(=\left(-25\cdot4\right)\cdot\left(-8\cdot125\right)\cdot\left(-17\right)=-1700000\)
29 tháng 5 2022
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
19 tháng 4 2017
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
T
28 tháng 10 2023
\(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{15}\right)+\dfrac{-6}{33}\right]+\dfrac{-3}{4}\)
\(=\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{5}{11}\cdot\dfrac{23}{13}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{115}{143}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\dfrac{-141}{143}-\dfrac{3}{4}\)
\(=-\dfrac{141}{104}-\dfrac{3}{4}\)
\(=-\dfrac{219}{104}\)
T
12 tháng 11 2023
\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)
\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)
\(=0\)
KV
9 tháng 12 2023
(2/5 + 2/7 - 2/11) : (3/7 - 3/11 + 3/5)
= 194/385 : 291/385
= 194/385 . 385/291
= 2/3
AH
Akai Haruma
Giáo viên
18 tháng 9 2018
Lời giải:
a)
\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)
b)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)
\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)
29 tháng 3
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
7 tháng 8
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(\begin{array}{l}A = \left( { - \dfrac{3}{{11}}} \right) + \dfrac{{11}}{8} - \dfrac{3}{8} + \left( { - \dfrac{8}{{11}}} \right)\\ = \left[ {\left( { - \dfrac{3}{{11}}} \right) + \left( { - \dfrac{8}{{11}}} \right)} \right] + \left( {\dfrac{{11}}{8} - \dfrac{3}{8}} \right)\\ = \dfrac{{ - 11}}{{11}} + \dfrac{8}{8} = - 1 + 1 = 0\end{array}\)